x + 2y - 1= -0
2x + y - 3 = 0
x - 2y - 3 = 0
2x + y - 1 = 0
Correct answer is D
Line PQ : x - 2y + 4 = 0
2y = x + 4 \(\implies y = \frac{x}{2} + 2 \)
Slope = \(\frac{1}{2}\)
Slope of the perpendicular line QR: \(\frac{-1}{\frac{1}{2}} = -2\)
Line QR: \(y = mx + b\)
\(y = -2x + b\)
Point of intersection: (1, -1)
\(-1 = -2(1) + b \implies b = -1 + 2 = 1\)
\(y = -2x + 1 \implies y + 2x - 1 = 0\)
\(QR: 2x + y - 1 = 0\)
\(\frac{1100}{3}\)km
\(\frac{2200}{3}\)km
\(\frac{22000}{3}\)km
\(\frac{1100}{3}\)km
Correct answer is C
Angular difference = 45° + 15° = 60°.
\(D = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 7000\)
= \(\frac{22000}{3} km\)
12o
56o
68o
78o
Correct answer is C
No explanation has been provided for this answer.
\(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\)
\(\begin{pmatrix} 2 & 7 \\ 4 & 17 \end{pmatrix}\)
\(\begin{pmatrix} 10 & 4 \\ 4 & 6 \end{pmatrix}\)
\(\begin{pmatrix} 4 & 3 \\ 10 & 9 \end{pmatrix}\)
Correct answer is A
\(\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\); y = \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\).
= \(\begin{pmatrix} 2 + 8 & 1 + 6 \\ 0 + 12 & 0 + 9 \end{pmatrix}\)
= \(\begin{pmatrix} 10 & 7 \\ 12 & 9 \end{pmatrix}\)
\(\begin{vmatrix} -2 & 1 & 1 \\ 2 & 1 & k \\1 & 3 & -1 \end{vmatrix}\) = 23
1
2
3
4
Correct answer is B
-2(-1 - 3k) - 1(-2 -k) + 1(6 - 1) = 23
7k = 14
k = 2